Can someone please explain bit masks and hexadecimal values for us to use with the protoboard.

Wed Oct 05, 2011 2:07 pm

MHTS

Guru

Joined: Sun Nov 15, 2009 5:46 amPosts: 1512

Re: Protoboard

Can you be more specific what you want to be explained?

Wed Oct 05, 2011 6:36 pm

maths222

Rookie

Joined: Wed Dec 01, 2010 5:15 pmPosts: 31

Re: Protoboard

I want to understand how to use bit masks to read digital values from the protoboard.

Mon Oct 24, 2011 8:53 am

MHTS

Guru

Joined: Sun Nov 15, 2009 5:46 amPosts: 1512

Re: Protoboard

To understand bit-wise operation, you must first understand hexadecimal number system. Computers only understand binary number system (base 2) because they only know 1's and 0's. The right most digit represents either a 0 or a 1, every subsequent digit to the left is an increment of the power of 2. So if the right most digit is 1, it represents one. If the next digit to the left is a 1, it represents 2. The next digit to the left represents 4, the next 8, the next 16 and so on. Therefore, the following decimal number can be translated to binary.

But binary numbers are not friendly to human because representing a typical number may require way too many digits. Hexadecimal system (base 16) is a good compromise because 16 is a power of 2 so it can be easily translated between binary and hexadecimal because it uses fewer digits. Every 4 binary bits can be combined to one hexadecimal digit. As the name implies a hexadecimal has 16 possible values (from 0 to 15). To represent the digit in a single character, we added letters (a-f) to represent 10-15. So the number 100 from the above example can be represented as:

To manipulate bits, you must understand the following bit-wise operators:- "&" This is to do a bit-wise "AND" of two binary numbers. Typically, it allows you to "clear" a bit in a variable. For example:

Code:

int num1 = 0x64; //decimal 100 or hex 64 or binary 0110 0100To clear the 3rd bit from the right, you will do the following:int num2 = num1 & 0xfb; //0110 0100 & 1111 1011When two bits "&" together, it will yield a "1" only if both operand bits are "1". So essentially, if you want to leave a bit alone "&" it with a "1". If you want to clear a bit, "&" it with a "0".

- "|" This is to do a bit-wise "OR" of two binary numbers. Typically, it allows you to "set" a bit in a variable. For example:

Code:

int num1 = 0x64; //decimal 100 or hex 64 or binary 0110 0100To set the 2nd bit from the right, you will do the following:int num2 = num1 | 0x02; //0110 0100 | 0000 0010When two bits "|" together, it will yield a "1" if any operand bits are "1". So essentially, if you want to leave a bit alone "|" it with a "0". If you want to set a bit, "|" it with a "1".

- "~" This is to do a bit-wise "NEGATE" of a binary number. Typically, it allows you to negate a value to form a bit-mask for clearing a bit. For example:

Code:

int num1 = 0x64; //decimal 100 or hex 64 or binary 0110 0100The third bit from the right is 0x04 (0000 0100)By negating it, you change all the 0's to 1's and all the 1's to 0's.~(0000 0100) = 1111 1011To clear the 3rd bit from the right, you will do the following:int num2 = num1 & ~0x04

- "^" This is to do a bit-wise "EXCLUSIVE-OR". Typically, it allows you to flip a bit. For example:

Code:

int num1 = 0x64; //decimal 100 or hex 64 or binary 0110 0100If we want to flip the bits of the lower nibble (0100), we do the following:int num2 = 0x64 ^ 0x0f; //0110 0100 ^ 0000 1111This will give you 0110 1011.When two bits "^" together, it will yield a "1" only if the two operand bits are different. It will yield a "0" if the operand bits are the same. So essentially, if you want to leave a bit alone "^" it with a "0". If you want to flip a bit, "^" it with a "1".

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